Lemma 20.7.2. Let $X$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. Let $U \subset X$ be an open subspace. Let $n > 0$ and let $\xi \in H^ n(U, \mathcal{F})$. Then there exists an open covering $U = \bigcup _{i\in I} U_ i$ such that $\xi |_{U_ i} = 0$ for all $i \in I$.

**Proof.**
Let $\mathcal{F} \to \mathcal{I}^\bullet $ be an injective resolution. Then

Pick an element $\tilde\xi \in \mathcal{I}^ n(U)$ representing the cohomology class in the presentation above. Since $\mathcal{I}^\bullet $ is an injective resolution of $\mathcal{F}$ and $n > 0$ we see that the complex $\mathcal{I}^\bullet $ is exact in degree $n$. Hence $\mathop{\mathrm{Im}}(\mathcal{I}^{n - 1} \to \mathcal{I}^ n) = \mathop{\mathrm{Ker}}(\mathcal{I}^ n \to \mathcal{I}^{n + 1})$ as sheaves. Since $\tilde\xi $ is a section of the kernel sheaf over $U$ we conclude there exists an open covering $U = \bigcup _{i \in I} U_ i$ such that $\tilde\xi |_{U_ i}$ is the image under $d$ of a section $\xi _ i \in \mathcal{I}^{n - 1}(U_ i)$. By our definition of the restriction $\xi |_{U_ i}$ as corresponding to the class of $\tilde\xi |_{U_ i}$ we conclude. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)